Largest IRG = 175.78125 MBytes"Calculating Largest IRG for 80% Data Tape

[nerdygazilio]

Homework Statement

Assume that a 2400 foot magnetic tape has recording density of 6400 Bpi. Data (logical) records are 100 bytes, and the memory buffer is 10,000 bytes. What is the largest IRG that will allow 80 percent of the tape to be data?

Homework Equations

and

The Attempt at a Solution

Tape length = 2400 ft = 2400×12 in=28800 inches
Recording density = 6400 bytes/in
Each record has 100 bytes = 64 inches
Memory buffer = 10,000 bytes
Largest IRG that allows 80% of tape to be data

Amount data = 80% of 2400 ft = 1920 ft consists of data.
Since each record has 100 bytes for 64 inches, then, 1 inch has 6400 bytes of data.
Total amount of data = 28800 inches × 6400 bytes = 184320000 bytes
= 184320000 bytes ÷ 220
= 175.78125 MBytes

 

[lewando]

What is the largest IRG that will allow 80 percent of the tape to be data?

This has not been answered. Is this what you are having trouble with?

Recording density = 6400 bytes/in
Each record has 100 bytes = 64 inches

If the density is 6400 bytes/in, then 6400 bytes = 1 in, and then 100 bytes should be much smaller than an inch.

Since each record has 100 bytes for 64 inches, then, 1 inch has 6400 bytes of data.

Somehow your bad premise resulted in a correct statement.

Total amount of data = 28800 inches × 6400 bytes = 184320000 bytes

True for 100% utilization (no IRG).

= 184320000 bytes ÷ 220

You have been pretty good at documenting your steps. Where does 220 come from?

= 175.78125 MBytes

How does 184,320,000/220 = 175,781,250? (I get the distinction you are using with "M", but still.../220?)

 

[nerdygazilio]

It is not two hundred and twenty(220)...it suppose to be 2²⁰ which gives the answer in megabytes same as multiplying 1024 by 1024..